Question

If $4cos{\left(36\right)}^o+cot\left(7{\left(\frac{1}{2}\right)}^{{}^{\circ }}\right)=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}$ $+\sqrt{n_4}+\sqrt{n_5}+\sqrt{n_6}$ , then the value of $\left(\frac{\sum _{i=1}^6{n}_i^2}{10}\right)$ is equal to

Answer 1

Answer: 9.1

$4cos{\left(36\right)}^o=4\left(\frac{\sqrt{5}+1}{4}\right)=\sqrt{5}+\sqrt{1}$
$cot7{\left(\frac{1}{2}\right)}^o=\frac{1+cos{\left(15\right)}^{{}^{\circ }}}{sin{\left(15\right)}^{{}^{\circ }}}=\frac{1+\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}$
$=\frac{\left(2\sqrt{2}+\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}$
$=\frac{2\sqrt{6}+2\sqrt{3}+2\sqrt{2}+4}{2}=\sqrt{6}+\sqrt{3}+\sqrt{2}+2=\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}$
$\Rightarrow 4cos{\left(36\right)}^o+cot\left(7{\left(\frac{1}{2}\right)}^o\right)=\sqrt{5}+\sqrt{1}+\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}$
$=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}$
$=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}+\sqrt{n_4}+\sqrt{n_5}+\sqrt{n_6}$
$\Rightarrow \sum _{i=1}^6{n}_i^2=1^2+2^2+3^3+4^2+5^2+6^2$